. Spherically symmetric charge distribution. Fig. 3-15 shows a spherical distribution of charge of radius R. The charge density at any points depends only on the distance of the point from the center and not on the direction, a condition called spherical symmetry. Find an expression for E for points (a) outside and (b) inside the charge distribution. Note that the object in Fig. 3-15 cannot be a conductor.
Solution.
"\\rho _0" ;
"R;"
"r;"
"a) E_{out}=\\dfrac{1}{4\\pi\\epsilon_0}\\dfrac{q_{tot}}{R^2};"
"q=\\int\\rho_0dV=\\rho_0(\\dfrac{4}{3}\\pi R^3);"
"E_{out}=\\dfrac{1}{4\\pi \\epsilon_0}\\dfrac{\\rho_0(\\dfrac{4}{3}\\pi R^3)}{R^2}=\\dfrac{\\rho_0 R}{3\\epsilon_0};"
"b)" "E_{in}=\\dfrac{1}{4\\pi\\epsilon_0}\\dfrac{q_{within}}{r^2};"
"q=\\int\\rho_0dV=\\rho_0(\\dfrac{4}{3}\\pi r^3);"
"E_{in}=\\dfrac{1}{4\\pi \\epsilon_0}\\dfrac{\\rho_0(\\dfrac{4}{3}\\pi r^3)}{r^2}=\\dfrac{\\rho_0 r}{3\\epsilon_0};"
Answer: "a)E_{out}=\\dfrac{\\rho_0 R}{3\\epsilon_0};"
"b)E_{in}=\\dfrac{\\rho_0 r}{3\\epsilon_0}."
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